我如何从十六进制字符串格式创建一个UIColor,如#00FF00?
当前回答
上面的几个解决方案涉及到一些不必要的nsstring使用。这个UIColor类扩展更简单更快:
+ colorWithHex:(UInt32)hex alpha:(CGFloat)alpha
{
return [UIColor colorWithRed:((hex & 0xFF0000) >> 16)/255.0
green:((hex & 0x00FF00) >> 8)/255.0
blue:( hex & 0x0000FF)/255.0
alpha:alpha];
}
简单来说:
return [UIColor colorWithHex:0x006400 alpha:1.0]; // HTML darkgreen
其他回答
extension UIColor
{
class func fromHexaString(hex:String) -> UIColor
{
let scanner = Scanner(string: hex)
scanner.scanLocation = 0
var rgbValue: UInt64 = 0
scanner.scanHexInt64(&rgbValue)
return UIColor(
red: CGFloat((rgbValue & 0xFF0000) >> 16) / 255.0,
green: CGFloat((rgbValue & 0x00FF00) >> 8) / 255.0,
blue: CGFloat(rgbValue & 0x0000FF) / 255.0,
alpha: CGFloat(1.0)
)
}
}
//you can call like this.
UIColor.fromHexaString(hex:3276b1)
extension UIColor {
convenience init(hexaString: String, alpha: CGFloat = 1) {
let chars = Array(hexaString.dropFirst())
self.init(red: .init(strtoul(String(chars[0...1]),nil,16))/255,
green: .init(strtoul(String(chars[2...3]),nil,16))/255,
blue: .init(strtoul(String(chars[4...5]),nil,16))/255,
alpha: alpha)}
}
用法:
let redColor = UIColor(hexaString: "#FF0000") // r 1,0 g 0,0 b 0,0 a 1,0
let transparentRed = UIColor(hexaString: "#FF0000", alpha: 0.5) // r 1,0 g 0,0 b 0,0 a 0,5
另一种选择是将六值转换为无符号整数,并从中提取相应的值:
extension UIColor {
convenience init(hexaString: String, alpha: CGFloat = 1) {
self.init(hexa: UInt(hexaString.dropFirst(), radix: 16) ?? 0, alpha: alpha)
}
convenience init(hexa: UInt, alpha: CGFloat = 1) {
self.init(red: .init((hexa & 0xff0000) >> 16) / 255,
green: .init((hexa & 0xff00 ) >> 8) / 255,
blue: .init( hexa & 0xff ) / 255,
alpha: alpha)
}
}
let purpleColor = UIColor(hexaString: "#FF00FF") // r 1,0 g 0,0 b 1,0 a 1,0
let transparentYellow = UIColor(hexaString: "#FFFF00", alpha: 0.5) // r 1,0 g 1,0 b 0,0 a 0,5
除了颜色,我还喜欢保证alpha,所以我自己写类别
+ (UIColor *) colorWithHex:(int)color {
float red = (color & 0xff000000) >> 24;
float green = (color & 0x00ff0000) >> 16;
float blue = (color & 0x0000ff00) >> 8;
float alpha = (color & 0x000000ff);
return [UIColor colorWithRed:red/255.0 green:green/255.0 blue:blue/255.0 alpha:alpha/255.0];
}
像这样很容易使用
[UIColor colorWithHex:0xFF0000FF]; //Red
[UIColor colorWithHex:0x00FF00FF]; //Green
[UIColor colorWithHex:0x00FF00FF]; //Blue
[UIColor colorWithHex:0x0000007F]; //transparent black
有cocoapod支持,这很好
https://github.com/mRs-/HexColors
// with hash
NSColor *colorWithHex = [NSColor colorWithHexString:@"#ff8942" alpha:1];
// wihtout hash
NSColor *secondColorWithHex = [NSColor colorWithHexString:@"ff8942" alpha:1];
// short handling
NSColor *shortColorWithHex = [NSColor colorWithHexString:@"fff" alpha:1]
Swift 2.0 - Xcode 7.2
为UIColor添加扩展。
文件-新建- Swift文件-命名。添加以下内容。
extension UIColor {
convenience init(hexString:String) {
let hexString:NSString = hexString.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet())
let scanner = NSScanner(string: hexString as String)
if (hexString.hasPrefix("#")) {
scanner.scanLocation = 1
}
var color:UInt32 = 0
scanner.scanHexInt(&color)
let mask = 0x000000FF
let r = Int(color >> 16) & mask
let g = Int(color >> 8) & mask
let b = Int(color) & mask
let red = CGFloat(r) / 255.0
let green = CGFloat(g) / 255.0
let blue = CGFloat(b) / 255.0
self.init(red:red, green:green, blue:blue, alpha:1)
}
func toHexString() -> String {
var r:CGFloat = 0
var g:CGFloat = 0
var b:CGFloat = 0
var a:CGFloat = 0
getRed(&r, green: &g, blue: &b, alpha: &a)
let rgb:Int = (Int)(r*255)<<16 | (Int)(g*255)<<8 | (Int)(b*255)<<0
return NSString(format:"#%06x", rgb) as String
}
}
用法:
Ex. Setting Button's color from hexCode.
override func viewWillAppear(animated: Bool) {
loginButton.tintColor = UIColor(hexString: " hex code here ")
}
Ex. Converting Button's current color to hex Code.
override func viewWillAppear(animated: Bool) {
let hexString = loginButton.tintColor.toHexString()
print("HEX STRING: \(hexString)")
}
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